# Thread: Complex Numbers Question, Involves Trig

1. ## Complex Numbers Question, Involves Trig

Hi, I'm looking for some help with complex numbers. I've attached a question below with the solution.

I understand where the route 41 is coming from but I am getting a different angle.

Part (b) I am also confused by.

Help is much appreciated.

2. ## Re: Complex Numbers Question, Involves Trig

Do you know that sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b)?

5cos(x)- 4 sin(x) cannot be "cos(a+ b)" for any a and b because "5" and "4" cannot be "sin(a)" and "cos(a)" for any a. sin(a) and cos(a) must be between -1 and 1 for all a and $\displaystyle sin^2(a)+ cos^2(a)= 1$. However, I note that $\displaystyle 5^2+ 4^2= 25+ 16= 41$ so that
$\displaystyle 5 cos(x)- 4 sin(x)= \sqrt{41}\left(\frac{5}{\sqrt{41}}cos(x)- \frac{4}{\sqrt{41}}sin(x)\right)$.

Now, what angle, a, has $\displaystyle sin(a)= \frac{5}{\sqrt{41}}$ and $\displaystyle cos(a)= -\frac{4}{\sqrt{41}}$?

3. ## Re: Complex Numbers Question, Involves Trig

For sin(a) I get 51.3 degrees and cos(a) - 38.7/321.3 degrees.

How does this relate to the answers given in the solution?

4. ## Re: Complex Numbers Question, Involves Trig

No, this is the same angle a for both. Sine is positive in the first and second quadrants while cosine is negative in the second and third quadrants. In order that sine be positive and cosine negative, the angle must be in the second quadrant- 180- 51.3= 128.7 degrees.
$\displaystyle sin(128.7)= 0.780= \frac{5}{\sqrt{41}}$ and $\displaystyle cos(128.7)= -0.63= -\frac{4}{\sqrt{41}}$

Now, remember the point was that $\displaystyle 5cos(x)- 4sin(x)= \sqrt{41}(sin(a) cos(x)+ cos(a)sin(x))= \sqrt{41}sin(x+ a)$

5. ## Re: Complex Numbers Question, Involves Trig

Right I get that, I remember the quadrants now. So from 128.7 how are the final two x values calculated?

Sorry but it's been a while since I've done this type of calculation.

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# cos 51.3degrees value

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